fixed typo
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@ -61,24 +61,24 @@ where $\mu = \bP X^{-1}$.
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\begin{theorem} % Theorem 3
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\begin{theorem} % Theorem 3
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\label{thm:lec10_3}
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\label{thm:lec10_3}
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Let $\bP \in M_1(\R)$ such that $\phi_\R \in L^1(\lambda)$.
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Let $\bP \in M_1(\R)$ such that $\phi_\bP \in L^1(\lambda)$.
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Then $\bP$ has a continuous probability density given by
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Then $\bP$ has a continuous probability density given by
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\[
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\[
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\R}(t) \dif t.
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f(x) = \frac{1}{2 \pi} \int_{\R} e^{-\i t x} \phi_{\bP}(t) \dif t.
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\]
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\]
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\end{theorem}
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\end{theorem}
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\begin{example}
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\begin{example}
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\begin{itemize}
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\begin{itemize}
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\item Let $\bP = \delta_{\{0\}}$.
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\item Let $\bP = \delta_{0}$.
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Then
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Then
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\[
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\[
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\phi_{\R}(t) = \int e^{\i t x} d \delta_0(x) = e^{\i t 0 } = 1
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\phi_{\bP}(t) = \int e^{\i t x} \dif \delta_0(x) = e^{\i t 0 } = 1
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\]
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\]
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\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
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\item Let $\bP = \frac{1}{2} \delta_1 + \frac{1}{2} \delta_{-1}$.
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Then
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Then
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\[
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\[
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\phi_{\R}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
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\phi_{\bP}(t) = \frac{1}{2} e^{\i t} + \frac{1}{2} e^{- \i t} = \cos(t)
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\]
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\]
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\end{itemize}
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\end{itemize}
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\end{example}
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\end{example}
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@ -88,20 +88,21 @@ where $\mu = \bP X^{-1}$.
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If $x_n \to x$, then $f(x_n) \to f(x)$.
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If $x_n \to x$, then $f(x_n) \to f(x)$.
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\end{claim}
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\end{claim}
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\begin{subproof}
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\begin{subproof}
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If $e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x } \phi(t) $ for all $t$.
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Suppose that
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$e^{-\i t x_n} \phi(t) \xrightarrow{n \to \infty} e^{-\i t x} \phi(t)$
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Then
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for all $t$.
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Since
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\[
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\[
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|e^{-\i t x} \phi(t)| \le |\phi(t)|
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|e^{-\i t x} \phi(t)| \le |\phi(t)|
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\]
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\]
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and $\phi \in L^1$, hence $f(x_n) \to f(x)$
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and $\phi \in L^1$,
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we get $f(x_n) \to f(x)$
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by the dominated convergence theorem.
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by the dominated convergence theorem.
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\end{subproof}
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\end{subproof}
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We'll show that for all $a < b$ we have
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We'll show that for all $a < b$ we have
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\[
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\[
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\bP\left( (a,b] \right) = \int_a^b (x) \dif x.\label{thm10_3eq1}
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\bP\left( (a,b] \right) = \int_a^b f(x) \dif x.\label{thm10_3eq1}
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\]
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\]
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Let $F$ be the distribution function of $\bP$.
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Let $F$ be the distribution function of $\bP$.
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It is enough to prove \autoref{thm10_3eq1}
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It is enough to prove \autoref{thm10_3eq1}
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@ -153,7 +154,7 @@ However, Fourier analysis is not only useful for continuous probability density
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Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
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Let $\phi$ be the characteristic function of $\bP \in M_1(\lambda)$.
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Then
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Then
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item $\phi(0) = 1$, $|\phi(t)| \le t$ and $\phi(\cdot )$ is continuous.
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\item $\phi(0) = 1$, $|\phi(1)| \le t$ and $\phi(\cdot )$ is continuous.
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\item $\phi$ is a \vocab{positive definite function},
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\item $\phi$ is a \vocab{positive definite function},
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i.e.~
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i.e.~
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\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
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\[\forall t_1,\ldots, t_n \in \R, (c_1,\ldots,c_n) \in \C^n ~ \sum_{j,k = 1}^n c_j \overline{c_k} \phi(t_j - t_k) \ge 0
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